DSE 2013 Answer Key is posted

[Granpa Simpson]

Yes Gujarati says so..:(...In that case it wud be better if we take this question to Akshay or Amit Sir I think..!!!

[phelps.phan]

Hi Kangkan/ Subhayu/anyone pls help me with the doubts I have. Thanks.

Q 1:The avg speed is 5 m/s. If she must have reached  5 m/s she must have run at 4 m/s at some point isn't it. Shouldn't the answer be a?

Q 38: Isn't all the locations between [0.2,0.7] pareto optimal? In this interval its not possible to make one person better off by making the other person worse off. As the distance increases for any one person, if the nuclear plant if moved in either direction .

Q 41: Are we implying that if exactly one has a positive probability, the other has 0 probability abd hence P(A).P(B)=0=P(AnB) ??

Q 50: Just one doubt here.The probability that X = b is not defined for a continious probability function right? Is it right to say it has positive probability.

[Granpa Simpson]

Q1) You can use Lagrange's mean Value theorem...velocity is dx/dt=f'(x)...now use LMVt within the interval and u will find dx/dt=5 atleast once..hence option d.
Q41) P(A intersection B)=P(A)*P(B) where nither P(A) nor P(B)=0 simultaneously according to the definition of independence of events..so P(B) cannot be zero neither can P(A) be so at the same time.
Q50) Point probability in a continuous distribution is zero but not undefined or indeterminate.

[kangkan]

Hi..[0.2,0.7] is not pareto optimal...for example let us consier x=2.0 if we shift it right to 1.99 both the agents are made better of..similarily the case with x=0.7

[phelps.phan]

Thanks Kangkan! Just to be clear, would you say the locations in the interval (0.2,0.7) instead of [0.2,0.7] is pareto optimal?

[kangkan]

i think so

[phelps.phan]

Thanks Subhayu.
For Q 41: Isn't P(AnB)=0 for disjoint events as AnB={}? If so, both P(A) and P(B) have positive probabilities would mean P(A).P(B) != P(AnB)=0 which means they cannot be independent(ans c). If A and B are independent besides being disjoint, only one of them has a positive probability, and the other 0. So either P(A) or P(B) is zero. I think this goes with the answer(option c) posted by Amit sir. This is what I've been trying to clarify.

[Granpa Simpson]

Yes that is the case..the condition is P(A) and P(B) cannot be zero simultaneously...this is because if that happens then the conditional probabilities will become undefined..so c is the correct answer.

[Akshay Jain]

The ques says that the var(error term) is not related to the variable x that is included in the model but to variable z dat is not included in the model...
if we use OLS in presence of hetroscedasticity den var(B2) vl be biased...the interpretation of this lies in the defination of heterscedasticity which states that var(error term) is related to a variable dat is included in the model....if var(error term) is related to sm other variable outside the model then var(B2) vl b unbiased.....
I hope this line of reasoning makes sence....

[Granpa Simpson]

Thanx a lot Akshay..gottit now...

[ragaland]

Akshay you have any idea what was the cut off last year??

[Akshay Jain]

Sry no idea about the cutoff....

[kangkan]

@akshay..if the variable Z affects the errot term in a sytematic way,then is nt model underfitted?in that case isnt the variance will be biased right?

[Akshay Jain]

yes u r absolutely right kangkan but here we hav no information that whether Z shud be included in the model or not.
if var(error) and Z are related den that doesnt mean that Z variable shud be included in the model or shud be excluded from the model. Gujarati text says that "it can cause bias" and not "It shud cause Bais"..
1 way to chek this is to compare adjusted R^2 or use F tests in restericted/unrestricted models..etc
the question is silent about these things...so best option is C
I hope it is clear now...

[kangkan]

alright..got it now..i hate the fact that dse asks us to select the best case....its so difficult to sure that you are not wrong

[Akshay Jain]

I would rather say that this is the only option and not the best option
this question tests you on 2 things...
1)the application of basic concept of heteroskedasticity
2)logical statements,ie, If "P is true then Q is true" this statement does not mean that "If Q is true then P is true".....
I hope m clear...:)

[Shefali]

kindly explain quest 6 plsss